The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y+4…

The vector equation of the plane through the line of intersection of the planes $x+y+z=1$ and $2 x+3 y+4 z=5$, which is perpendicular to the plane $x-y+z=0$, is
  1. $\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})=2$
  2. $\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{k}})+2=0$
  3. $\bar{r} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{k}})=2$
  4. $\quad \overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$

Solution

The equation of the required plane $\begin{aligned} & (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\ & \Rightarrow(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z-1-5 \lambda=0 \end{aligned}$
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the d.r.s. of the required plane. $\therefore \quad$ From (i), $a=1+2 \lambda, b=1+3 \lambda, \mathrm{c}=1+4 \lambda$ The required plane is perpendicular to $x-y+z=0$ $\begin{aligned} \therefore \quad & a-b+c=0 \\ & \Rightarrow 1+2 \lambda-(1+3 \lambda)+1+4 \lambda=0 \\ & \Rightarrow 1+3 \lambda=0 \\ & \Rightarrow \lambda=-\frac{1}{3} \end{aligned}$
Substituting $\lambda=-\frac{1}{3}$ in (i), we get $\begin{aligned} & \left(1-\frac{2}{3}\right) x+\left(1-\frac{3}{3}\right) y+\left(1-\frac{4}{3}\right) z-1+\frac{5}{3}=0 \\ & \Rightarrow x-z+2=0 \end{aligned}$
Its vector equation is $\overline{\mathrm{r}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{k}})+2=0$

Asked in: MHT CET 2024 (15 May Shift 2)

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